KnowledgeBoat Logo
|

Mathematics

Prove the following identity:

(1+sin⁡θ)2+(1−sin⁡θ)22cos⁡2θ=sec⁡2θ+tan⁡2θ\dfrac{(1 + \sin \theta)^2 + (1 - \sin \theta)^2}{2 \cos^2 \theta} = \sec^2 \theta + \tan^2 \theta

Trigonometric Identities

2 Likes

Answer

Solving L.H.S. of the above equation :

⇒(1+2sin⁡θ+sin⁡2θ)+(1−2sin⁡θ+sin⁡2θ)2cos⁡2θ⇒2+2sin⁡2θ2cos⁡2θ⇒2(1+sin⁡2θ)2cos⁡2θ⇒(1+sin⁡2θ)cos⁡2θ⇒1cos⁡2θ+sin⁡2θcos⁡2θ⇒sec⁡2θ+tan⁡2θ.\Rightarrow \dfrac{(1 + 2\sin \theta + \sin^2 \theta) + (1 - 2\sin \theta + \sin^2 \theta)}{2 \cos^2 \theta} \\[1em] \Rightarrow \dfrac{2 + 2 \sin^2 \theta}{2 \cos^2 \theta} \\[1em] \Rightarrow \dfrac{2(1 + \sin^2 \theta)}{2 \cos^2 \theta} \\[1em] \Rightarrow \dfrac{(1 + \sin^2 \theta)}{\cos^2 \theta} \\[1em] \Rightarrow \dfrac{1}{\cos^2 \theta} + \dfrac{\sin^2 \theta}{\cos^2 \theta} \\[1em] \Rightarrow \sec^2 \theta + \tan^2 \theta.

Since, L.H.S. = R.H.S.

Hence, proved that (1+sin⁡θ)2+(1−sin⁡θ)22cos⁡2θ=sec⁡2θ+tan⁡2θ\dfrac{(1 + \sin \theta)^2 + (1 - \sin \theta)^2}{2 \cos^2 \theta} = \sec^2 \theta + \tan^2 \theta.

Answered By

3 Likes

Related Questions